1.2 KiB
1.2 KiB
| id | title | challengeType | forumTopicId | dashedName |
|---|---|---|---|---|
| 5900f3e61000cf542c50fef9 | Problem 122: Efficient exponentiation | 5 | 301749 | problem-122-efficient-exponentiation |
--description--
The most naive way of computing n^{15} requires fourteen multiplications:
n × n × \ldots × n = n^{15}But using a "binary" method you can compute it in six multiplications:
\begin{align}
& n × n = n^2\\\\
& n^2 × n^2 = n^4\\\\
& n^4 × n^4 = n^8\\\\
& n^8 × n^4 = n^{12}\\\\
& n^{12} × n^2 = n^{14}\\\\
& n^{14} × n = n^{15}
\end{align}$$
However it is yet possible to compute it in only five multiplications:
$$\begin{align}
& n × n = n^2\\\\
& n^2 × n = n^3\\\\
& n^3 × n^3 = n^6\\\\
& n^6 × n^6 = n^{12}\\\\
& n^{12} × n^3 = n^{15}
\end{align}$$
We shall define $m(k)$ to be the minimum number of multiplications to compute $n^k$; for example $m(15) = 5$.
For $1 ≤ k ≤ 200$, find $\sum{m(k)}$.
# --hints--
`efficientExponentation()` should return `1582`.
```js
assert.strictEqual(efficientExponentation(), 1582);
```
# --seed--
## --seed-contents--
```js
function efficientExponentation() {
return true;
}
efficientExponentation();
```
# --solutions--
```js
// solution required
```